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Chapter 19: Problem 89

A \(0.10-\mathrm{cm}^{3}\) sample of a solution containing a radioactive nuclide $\left(5.0 \times 10^{3} \text { counts per minute per milliter) is injected }\right.\( into a rat. Several minutes later 1.0 \)\mathrm{cm}^{3}$ of blood is removed. The blood shows 48 counts per minute of radioactivity. Calculate the volume of blood in the rat. What assumptions must be made in performing this calculation?

Short Answer

Expert verified
The volume of blood in the rat is approximately 10.42 cm³, assuming that the radioactive nuclide is evenly distributed in the rat's blood and the concentration of radioactivity in the blood sample is representative of the concentration in the rat's entire blood volume.

Step by step solution

01

Calculate the total counts of radioactivity

The problem states that the injected radioactive nuclide concentration is 5.0x10^3 counts per minute per milliliter (cpm/ml) and the sample volume is 0.10 cm³. To find the total counts of radioactivity, we'll multiply both values: Total counts of radioactivity \(= \text{nuclide concentration} \times \text{sample volume}\) Total counts of radioactivity \(= (5.0 \times 10^{3} \,\text{cpm/ml}) \times 0.10\, \mathrm{cm}^{3}\) #Step 2: Calculate the total counts of radioactivity injected into the rat#
02

Calculate the total counts

Now, we can calculate the total counts of radioactivity: Total counts of radioactivity \(= (5.0 \times 10^{3} \,\text{cpm/ml}) \times 0.10\, \mathrm{cm}^{3} = 500\, \mathrm{cpm}\) #Step 3: Find the concentration of the radioactive nuclide in the rat's blood#
03

Determine the concentration in the rat's blood

The blood sample has a volume of 1.0 cm³ and shows 48 counts per minute of radioactivity. Thus, the concentration of the radioactive nuclide in the rat's blood can be calculated as: Blood concentration \(= \cfrac{\text{counts per minute in blood sample}}{\text{volume of blood sample}}\) Blood concentration \(= \cfrac{48\, \mathrm{cpm}}{1.0\,\mathrm{cm^{3}}}\) Blood concentration \(= 48\, \mathrm{cpm/cm^{3}}\) #Step 4: Calculate the volume of blood in the rat#
04

Calculate the blood volume

Finally, to find the volume of blood in the rat, we'll use the total counts of radioactivity and the concentration of the radioactive nuclide in the rat's blood: Blood volume \(= \cfrac{\text{total counts of radioactivity}}{\text{blood concentration}}\) Blood volume \(= \cfrac{500\, \mathrm{cpm}}{48\, \mathrm{cpm/cm^{3}}}\) Blood volume \(= 10.42\, \mathrm{cm^{3}}\) #Step 5: State the assumptions made in this calculation#
05

List the assumptions

In performing this calculation, we made the following assumptions: 1. The radioactive nuclide is evenly distributed in the rat's blood. 2. The concentration of radioactivity in the blood sample is representative of the concentration in the rat's entire blood volume. The volume of blood in the rat is approximately 10.42 cm³.

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Most popular questions from this chapter

Iodine- 131 is used in the diagnosis and treatment of thyroid disease and has a half-life of 8.0 days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing $10 . \mu \mathrm{g}^{131 \mathrm{I}}\( how long will it take for the amount of \)^{131} \mathrm{I}$ to decrease to 1\(/ 100\) of the original amount?

To determine the \(K_{\mathrm{sp}}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2},\) a chemist obtained a solid sample of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in which some of the iodine is present as radioactive 131 \(\mathrm{I}\) . The count rate of the \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) sample is \(5.0 \times 10^{11}\) counts per minute per mole of I. An excess amount of $\mathrm{Hg}_{2} \mathrm{I}_{2}(s)$ is placed into some water, and the solid is allowed to come to equilibrium with its respective ions. A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) $$ \mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \qquad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$

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